Question

When a photon is emitted by a hydrogen atom, the photon carries a momentum with it. (a) Calculate the momentum carries by the photon when a hydrogen atom emits light of wavelength 656.3 nm. (b) With what speed does the atom recoil during this transition? Take the mass of the hydrogen atom = 1.67 × 10−27 kg. (c) Find the kinetic energy of recoil of the atom.

Answer 1

Explanation:

When a photon is emitted by a hydrogen atom, the photon carries a momentum with it. (a) Calculate the momentum carries by the photon when a hydrogen atom emits light of wavelength 656.3 nm. (b) With what speed does the atom recoil during this transition? Take the mass of the hydrogen atom = 1.67 × 10−27 kg. (c) Find the kinetic energy of recoil of the atom.

Answer 2

(a) The momentum carries by the  photon when a hydrogen atom emits light of wavelength 656.3 n=m is $p = 1\times 10^{-27}$ kg-m/s

(b) 0.6 m/s  is the speed does the atom recoil during the transition

(c) $K = 1.9 \times 10^{-9}$ eV is the Kinetic energy of the recoil of the atom

Explanation:

It is given that the light’s wavelength emitted by hydrogen, λ = 656.3 n-m

Hydrogen atom’s mass, m = 1.67 × 10−27 kg

(a) Momentum

P is shown as P = hλ

where, λ = light’s wavelength and h = constant of Planck  

Therefore, $p = \frac{6.63 \times 10-34}{656.3 \times 10-9}$

 $p = 1\times 10^{-27}$ kg-m/s                        

(b) Momentum, p = m×v  

where, m = hydrogen atom’s mass; v = Speed of atom

Therefore, p= (1.67 × 10−27)× v

Substitute the value of p in above equation

we get, v = 0.6 m/s

(c) Kinetic energy

Atom’s recoil is shown as $K =\frac{1}{2} mv^2$      

where, m = atom’s mass, v = atom’s velocity  

Therefore,  $K = \frac{1}{2} \times 1.67 \times 10^{-27} \times 0.6^{2}$

$K = 1.9 \times 10^{-9}$ eV